How to get the moles and molarity? Part C. Determination of UnknownBalanced chemical equation:Hcl + Na OH -o Nacl t Hg0Unknown number NATrial 1Trial 2Trial 3Volume of unknown HCltitrated10.00mLNANAInitial burette reading0.90mLFinal burette reading30.90mLVolume NaOH added20 00ML50.00mLXO.144 m=43.2mmoles NaOH added4.335m(Use average molarity from part Band multiply it by the volume you added)04335 mmoles HCl4.335m04335m10.00 mLmolarity of HClActual Molarity (from instructor)0.3150Mpercentage error

We have to find moles, molarity of HCl

Answered: O,4335 m moles HCI 4.335m =0.4335m…

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O,4335 m moles HCI 4.335m =0.4335m 10.00ML molarity of HCI

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

How to get the moles and molarity?

Part C. Determination of Unknown
Balanced chemical equation:
Hcl + Na OH -o Nacl t Hg0
Unknown number NA
Trial 1
Trial 2
Trial 3
Volume of unknown HCl
titrated
10.00mL
NA
NA
Initial burette reading
0.90mL
Final burette reading
30.90mL
Volume NaOH added
20 00ML
50.00mLXO.144 m=43.2m
moles NaOH added
4.335m
(Use average molarity from part B
and multiply it by the volume you added)
04335 m
moles HCl
4.335m
04335m
10.00 mL
molarity of HCl
Actual Molarity (from instructor)
0.3150M
percentage error

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